Let $X := \mathbb R^d$, $\lambda^d$ be the $d$-dimensional Lebesgue measure on $X$, and $f:X \to \mathbb R$ convex. Then [there is](https://math.stackexchange.com/questions/4483452/is-it-true-that-if-f-mathbb-rd-to-mathbb-r-is-continuous-then-the-differ) a Borel set $N \subset X$ such that $\lambda^d (N) = 0$ and $f$ is differentiable on $A := X \setminus N$. Clearly, $A$ is Borel measurable. It follows that the gradient $\nabla f: A \to X$ of $f$ is well-defined.

- If $f$ is differentiable on $X$, [then](https://math.stackexchange.com/questions/4482935/if-all-partial-derivatives-exist-at-each-point-of-a-then-f-in-mathcal-c1) $A = X$ and $f \in \mathcal C_1 (X)$.

- Clearly, the Borel $\sigma$-algebra $\mathcal B(A)$ is a subset of the Borel $\sigma$-algebra $\mathcal B(X)$, i.e., $\mathcal B(A) \subset \mathcal B(X)$. 

**My question:** Is the map $\nabla f$ measurable?